AOS3 Topic 9: Optimisation Problems

Optimization problems are a fundamental aspect of mathematics and various applied sciences, where the goal is to find the best solution from a set of possible choices. These problems typically involve maximizing or minimizing a function under certain constraints.

Introduction to Optimization

Optimization is the process of finding the most efficient, cost-effective, or highest achievable performance solution to a problem, given certain constraints and conditions. It plays a crucial role in fields such as economics, engineering, operations research, and more.

Types of Optimization Problems

Unconstrained Optimization: Involves finding the maximum or minimum of a function without any restrictions. The problem is solved by setting the derivative of the function to zero to find critical points, which are then tested to determine if they are maxima or minima.

Constrained Optimization: Involves optimizing a function subject to certain constraints, such as inequalities or equalities. Common methods to solve these problems include the Lagrange multiplier method and the Karush-Kuhn-Tucker (KKT) conditions.

Linear Optimization (Linear Programming): A special case of optimization where both the objective function and the constraints are linear. This is widely used in business and economics for resource allocation, production planning, and scheduling.

Nonlinear Optimization: Deals with problems where the objective function or the constraints are nonlinear. These problems are more complex and require specialized algorithms for their solution, such as gradient descent, Newton's method, or evolutionary algorithms.

Solving Optimization Problems

Define the Objective Function: Clearly state what needs to be maximized or minimized. This could be a cost function, profit function, distance, time, etc.

Determine the Constraints: Identify the constraints that the solution must satisfy. These can be equations or inequalities that limit the feasible region of possible solutions.

Use Analytical or Numerical Methods: Depending on the nature of the problem, apply appropriate mathematical techniques or algorithms to find the optimal solution. Analytical methods involve calculus and algebra, while numerical methods use computational approaches.

Interpret the Solution: Once the optimal solution is found, it’s crucial to interpret it in the context of the problem to ensure it makes practical sense.

Guideline for Solving Optimization Problems

Identify what is to be maximized or minimized and what the constraints are.

Draw a diagram (if appropriate) and label it.

Decide what the variables are and in what units their values are being measured. For example, A for area in square meters, r for radius in inches, C for cost in Euros. In other words, if the problem does not introduce these variables, you need to do so.

Write a formula for the function that is to be maximized or minimized.

Use the given constraint to express the formula from Step 4 in terms of a single variable, namely something like f(x) (or A(x), C(x), etc., whatever name is appropriate). Then identify the domain of this function, which is typically [a, b] or (a, b).

Find the critical points of f. Compare all critical values and endpoints (or perhaps lim_x→a+ f(x) and lim_x→a− f(x) or curve sketching if the interval is open) to determine the absolute extrema of f.

Provide your solution meaningfully, which includes unit(s).

Applications of Optimization

Optimization is used in a variety of fields:

Engineering: Design of efficient systems, such as minimizing material use while maintaining strength in structural engineering.
Economics: Maximizing profit or utility functions while considering resource constraints.
Operations Research: Solving problems related to logistics, supply chain management, and production planning.
Machine Learning: Training models by optimizing a loss function to improve accuracy.

Example 1

Maximizing Profit

You want to sell a certain number n of items in order to maximize your profit. Market research tells you that if you set the price at $1.50, you will be able to sell 5000 items, and for every 10 cents you lower the price below $1.50, you will be able to sell another 1000 items. Suppose that your fixed costs ("start-up costs") total $2000, and the per item cost of production ("marginal cost") is $0.50.

Problem: Find the price to set per item and the number of items sold in order to maximize profit, and also determine the maximum profit you can get.

Solution:

The first step is to convert the problem into a function maximization problem. Since we want to maximize profit by setting the price per item, we should look for a function P(x) representing the profit when the price per item is x.

Profit is revenue minus costs, and revenue is the number of items sold times the price per item, so we get:

P = n * x - 2000 - 0.50 * n.

The number of items sold is itself a function of x,

n = 5000 + 1000 * (1.5 - x) / 0.10,

because (1.5 - x) / 0.10 is the number of multiples of 10 cents that the price is below $1.50. Now we substitute for n in the profit function:

P(x) = (5000 + 1000 * (1.5 - x) / 0.10) * x - 2000 - 0.5 * (5000 + 1000 * (1.5 - x) / 0.10)

= -10000x² + 25000x - 12000

We want to know the maximum value of this function when x is between 0 and 1.5. The derivative is:

P'(x) = -20000x + 25000,

which is zero when x = 1.25.

Since P''(x) = -20000 < 0, there must be a relative maximum at x = 1.25, and since this is the only critical value, it must be a global maximum as well. (Alternatively, we could compute P(0) = -12000, P(1.25) = 3625, and P(1.5) = 3000 and note that P(1.25) is the maximum of these.) Thus, the maximum profit is $3625, attained when we set the price at $1.25 and sell 7500 items.

Example 2

Minimize Average Cost

A manufacturer determines that the daily average cost of producing q units is:

$ \overline{C}(q) = 0.0001q^2 - 0.08q + 65 + \frac{5000}{q} $, $ q > 0 $

Problem: Determine the number of units produced per day which minimizes the average cost.

Solution

We first note that the domain of the function $ \overline{C} $ is the open interval $ (0, \infty) $. Calculate:

$ \overline{C}'(q) = 0.0002q - 0.08 - \frac{5000}{q^2} $

Then solving $ \overline{C}'(q) = 0 $ gives $ q = 500 $, which is the only critical point of $ \overline{C} $. Next:

$ \overline{C}''(q) = 0.0002 + \frac{10,000}{q^3} $

And so:

$ \overline{C}''(500) = 0.0002 + \frac{10,000}{500^3} > 0 $

Therefore, by the Second Derivative Test, $ q = 500 $ is a relative minimum of $ \overline{C} $.

Furthermore, $ \overline{C} $ is concave upward everywhere, so the relative minimum of $ \overline{C} $ must be the absolute minimum of $ \overline{C} $. The minimum cost is thus:

$ \overline{C}(500) = 0.0001(500)^2 - 0.08(500) + 65 + \frac{5000}{500} = 60 $

or $60 per unit. The sketch of $ \overline{C} $ follows.

Example 3

Minimize Travel Time

Suppose you want to reach a point A that is located across the sand from a nearby road (see diagram below). Suppose that the road is straight, and b is the distance from A to the closest point C on the road. Let v be your speed on the road, and let w, which is less than v, be your speed on the sand. Right now you are at the point D, which is a distance a from C. At what point B should you turn off the road and head across the sand in order to minimize your travel time to A?

Solution

Let x be the distance short of C where you turn off, i.e., the distance from B to C. We want to minimize the total travel time. Recall that when traveling at constant velocity, time is distance divided by velocity.

You travel the distance $ \overline{DB} $ at speed v, and then the distance $ \overline{BA} $ at speed w. Since $ \overline{DB} = a - x $ and, by the Pythagorean Theorem, $ \overline{BA} = \sqrt{x^2 + b^2} $, the total time for the trip is:

$ f(x) = \frac{a - x}{v} + \frac{\sqrt{x^2 + b^2}}{w} $

We want to find the minimum value of f when x is between 0 and a. As usual, we set $ f'(x) = 0 $ and solve for x:

$ 0 = f'(x) = -\frac{1}{v} + \frac{x}{w\sqrt{x^2 + b^2}} $

Solving this gives:

$ \frac{v\sqrt{x^2 + b^2}}{w} = x $

Which simplifies to:

$ x = \frac{wb}{\sqrt{v^2 - w^2}} $

Notice that a does not appear in the last expression, but a is not irrelevant, since we are interested only in critical values that are in $ [0, a] $. If $ \frac{wb}{\sqrt{v^2 - w^2}} $ is within this interval, we can use the second derivative to test it:

$ f''(x) = \frac{b^2}{(x^2 + b^2)^{3/2}w} $

Since this is always positive, there is a relative minimum at the critical point, and so it is a global minimum as well.

If the critical value is not in $ [0, a] $, it is larger than a. In this case, the minimum must occur at one of the endpoints. We can compute:

$ f(0) = \frac{a}{v} + \frac{b}{w} $ and $ f(a) = \frac{\sqrt{a^2 + b^2}}{w} $

But it is difficult to determine which of these is smaller by direct comparison. If, as is likely in practice, we know the values of v, w, a, and b, then it is easy to determine this. With a little cleverness, however, we can determine the minimum in general. We have seen that $ f''(x) $ is always positive, so the derivative $ f'(x) $ is always increasing. We know that at $ \frac{wb}{\sqrt{v^2 - w^2}} $ the derivative is zero, so for values of x less than that critical value, the derivative is negative. This means that $ f(0) > f(a) $, so the minimum occurs when $ x = a $.

So the upshot is this: If you start farther away from C than $ \frac{wb}{\sqrt{v^2 - w^2}} $, then you always want to cut across the sand when you are a distance $ \frac{wb}{\sqrt{v^2 - w^2}} $ from point C. If you start closer than this to C, you should cut directly across the sand.

Exercise 1

Practice Makes Perfect

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